Cameroon GCE A-Level Pure Mathematics with Statistics common mistakes
Cameroon GCE A-Level Pure Mathematics with Statistics · Understand the traps and how to avoid them.
When solving an inequality such as (x-2)(x+5) < 0, the candidate writes x < 2 and x < -5 (or x > 2 and x > -5) and presents the intersection or union of the two half-lines as the answer, instead of testing the sign of the product on each interval and concluding -5 < x < 2.
The candidate treats an inequality as two independent inequalities to be solved separately, carrying over the habit from simultaneous equations where each factor is set to zero. The sign interaction between factors, which is the whole content of a quadratic inequality, is never analysed, and sketching the parabola is skipped because it feels like extra work.
What to do instead
Adopt a fixed two-step routine: solve the corresponding equation to get the critical values, then draw a quick sign table or sketch of y = (x-2)(x+5) marking where the curve crosses the axis, and read off the interval where the curve lies below or above the axis. Always test one interior point (here x = 0 gives -10 < 0, so the region between the roots is the solution) and one exterior point before writing the final answer in interval notation.
When finding the sum to infinity of a geometric series, the candidate uses S = a/(1-r) without first establishing |r| < 1, and applies it to a series such as 3 + 6 + 12 + ... or to a series where r was computed with a sign error, producing a negative or nonsensical sum instead of stating that the series diverges.
The formula is memorised as a one-line recipe and the convergence condition is treated as a footnote. Sign errors in r are common when the common ratio is negative, and the candidate does not check the computed r against the terms of the series, so a divergent series slips through the formula.
What to do instead
Before every use, write the condition |r| < 1 explicitly and verify it with the numbers: compute r by dividing two consecutive terms and check it reproduces a third term. If |r| >= 1, state that the series diverges and the sum to infinity does not exist. For negative ratios, keep the sign inside the bracket, e.g. S = a/(1-(-1/3)) = a/(4/3), rather than dropping the double negative.
When finding the equation of a tangent or normal to a curve at a given point, the candidate substitutes the x-coordinate into the original equation to get the gradient, or uses the derivative as the gradient of the normal without taking the negative reciprocal, and then writes the line as y = m x + c using the wrong m.
The roles of the function and its derivative are conflated under exam pressure: both involve 'putting x into a formula'. The negative reciprocal rule for perpendicular lines is recalled as 'flip it' but the sign change is dropped, and the candidate does not substitute the point back into the final line to check it.
What to do instead
Write three labelled lines before touching the algebra: (i) point (x1, y1) with y1 from the original equation, (ii) gradient m = f'(x1) from the derivative, (iii) for a normal m_normal = -1/m. Then use y - y1 = m(x - x1) rather than hunting for c. Finish by substituting the point into the final equation as a ten-second check, and confirm the normal's gradient is negative reciprocal of the tangent's.
When integrating by substitution or by parts, the candidate changes the variable in the integrand but leaves the limits in terms of the old variable, or omits the constant of integration on an indefinite integral, or drops the dx factor during the substitution.
The mechanics of the substitution absorb all the attention, and the limits sit at the ends of the integral sign where they are easy to forget once the integrand has been rewritten. The constant of integration is treated as cosmetic because in definite integrals it cancels, so the habit of omitting it carries over.
What to do instead
Use a written checklist for every substitution: differentiate the substitution to get dx in terms of du, convert the limits explicitly (write 'when x = 0, u = 1' etc.), and only then integrate. For indefinite integrals, make '+ C' the last thing written on the line before moving on. For integration by parts, write u, du, dv, v in a small table first so the signs and the remaining integral are unambiguous.
When answering a probability question involving 'at least one' or a union of events, the candidate adds the individual probabilities, e.g. P(A or B) = P(A) + P(B) without subtracting P(A and B), or computes 'at least one' as n times the single-event probability, exceeding 1 without noticing.
The addition rule is remembered in its mutually exclusive form because that is the version most often practised, and the candidate does not check whether the events can occur together. 'At least one' is read as 'one', collapsing a complement problem into a single trial.
What to do instead
For any union, first ask whether the events are mutually exclusive; if not, use P(A or B) = P(A) + P(B) - P(A and B), and draw a two-circle Venn diagram or two-way table to see the overlap. For 'at least one' in n independent trials, always use 1 - P(none) = 1 - (1-p)^n. Treat any probability outside [0,1] as a signal that the method is wrong, not a rounding issue.
When reading a cumulative frequency curve or histogram, the candidate reads the median by taking the halfway point on the vertical axis and projecting to the curve but then reports the vertical value, or in a histogram with unequal class widths uses frequency rather than frequency density on the vertical axis, so bar heights and areas are misread.
The axes of a cumulative frequency graph are swapped relative to the usual 'x then y' reading habit, and the candidate projects to the wrong axis. In histograms, the area-equals-frequency principle is learned as a slogan but not applied when class widths differ, so the taller bar is assumed to represent the larger frequency.
What to do instead
For cumulative frequency: locate n/2 on the vertical (cumulative frequency) axis, move horizontally to the curve, then drop vertically to the horizontal axis and read the variable there; label the axes on the diagram before starting. For histograms: compute frequency density = frequency / class width for every class before drawing or reading, and remember that frequency is the area of the bar, so compare areas, not heights, when class widths differ.
In a hypothesis test for a population mean with a small sample, the candidate uses the normal distribution and the given sigma instead of the t-distribution with n-1 degrees of freedom and the sample standard deviation, or states the conclusion as 'accept H0' rather than 'do not reject H0', and reports the p-value against the wrong tail for a one-sided test.
The normal test is the first and most practised case, and the switch to t for small samples is triggered by sample size only if the candidate consciously checks it. The language 'accept H0' comes from everyday speech, and the tail is chosen by habit rather than by reading the direction of the alternative hypothesis.
What to do instead
Write the test setup as four explicit lines: H0, H1 (with direction), significance level, and the test statistic with its distribution justified by sample size and whether sigma is known. Use t with n-1 degrees of freedom whenever sigma is unknown and n is small. Halve the two-tailed p-value only for a one-sided test, and phrase the conclusion as 'there is insufficient evidence to reject H0 at the 5% level', never 'H0 is accepted'.
When multiplying or finding the inverse of 2x2 or 3x3 matrices, the candidate computes the determinant as ad - bc but then inverts using (1/det) times the original matrix without swapping a and d and negating b and c, or multiplies matrices in the wrong order (AB vs BA) when solving a system.
The inverse formula is a visual pattern that is easy to half-remember: the determinant step is correct, but the cofactor rearrangement is recalled only vaguely. Matrix multiplication is assumed to be commutative by analogy with number arithmetic, so the order of the product is not policed.
What to do instead
Write the inverse formula in full each time: for M = [[a,b],[c,d]], M^-1 = (1/(ad-bc)) [[d,-b],[-c,a]], and check by multiplying M M^-1 to get I. For solving AX = B, premultiply both sides by A^-1 to get X = A^-1 B, and never write B A^-1. For 3x3 matrices, find the determinant by expansion along a row with alternating signs written above the entries.
When working with complex numbers, the candidate writes the modulus of a sum as the sum of moduli (|z1 + z2| = |z1| + |z2|), or forgets to multiply by the conjugate when rationalising a quotient such as (3+2i)/(1-i), leaving an imaginary part in the denominator, or loses the sign when converting between Cartesian and polar form in the wrong quadrant.
Modulus behaves like absolute value for real numbers, so the additive property is assumed to carry over. Rationalising surds is familiar from earlier work but the complex conjugate is a new step, and the quadrant of the argument depends on the signs of both parts, which are often read from a sketch made too hastily.
What to do instead
Treat |z| as a distance: compute it from sqrt(a^2 + b^2) and never split across a sum. For quotients, always multiply numerator and denominator by the conjugate of the denominator and show the denominator becoming a^2 + b^2, a real number. For polar form, plot the point, determine the quadrant first, then compute the argument with atan(b/a) and adjust by pi when a < 0, stating the principal value.
When differentiating a composite function such as sin(3x^2) or (2x+1)^5, the candidate differentiates the outer function only and writes 5(2x+1)^4 or cos(3x^2), omitting the chain rule factor, or differentiates a product such as x^2 e^x as 2x e^x, omitting the product rule term.
The outer derivative is the salient step and the inner derivative feels like a formality, especially when the inner function is linear and its derivative is a constant that is easy to overlook. For products, the two factors are differentiated separately and combined by habit rather than by rule.
What to do instead
Name the inner function explicitly: let u = 3x^2, so y = sin u, then dy/dx = cos u times du/dx = 6x cos(3x^2). For products, write the rule u'v + uv' at the top of the working and fill the four pieces in before simplifying. As a check, differentiate a simple case numerically or verify that the degree or period of the answer is consistent with the original function.